How can I find the equation of one parabola given the equation of another parabola?

Question

The equation of the other parabola has to follow the form:

$4p(x – h) = (y – k)^2$

because it is a sideways parabola. I can see that the vertex is at (-4.5,18)

So then the equation would be $4p(x+4.5) = (y-18)^2$

Now I can just plug in a value like (4,5) for x and y in that equation and solve for p.

If I couldn't see the values of the vertex on the graph, how would I solve this problem?

Answer 1

Assuming the axis of the solid parabola is parallel to the $x$-axis...

Plug in the points. \begin{align*} 4 p(4 - h) &= (5 - k)^2 \text{,} \\ 4 p(-2 - h) &= (11 - k)^2 \text{, and} \\ 4 p(-4 - h) &= (21 - k)^2 \text{.} \end{align*} Then solve for $h$, $p$, and $k$. You'll discover $h \neq -4.5$.

(As a hint for solving: note that subtracting any of these from any of the others cancels the $k^2$ and the $-4ph$. In this way you can convert to three linear equations with two unknowns. Solve them, then plug back into any of the above to get the third.)

Answer 2

There are an infinite number of parabolas that go through those three points. (Most of them have an axis that is not parallel to either of the coordinate axes.) So if you don't have the clue from the graph that the axis of the other parabola is parallel to the $x$-axis, you can't solve the problem.

In fact, although it's pretty clear what is intended, strictly speaking the question should have told you that the other parabola has axis parallel to the $x$-axis (a "sideways" parabola).

By the way, how do you "see" that the vertex is at $(-4.5,18)$? Visual inspection is not good enough! (And it seems from Eric Towers' answer that you are wrong about this.)

Answer 3

The best solution is to write three equations using the three given points:

Eq. 1: $4p(4-h)=(5−k)^2$

Eq. 2: $4p(-2-h)=(11−k)^2$

Eq. 3: $4p(-4-h)=(21−k)^2$

And use substitution to solve for p, k, and h. To begin:

From Eq. 1: $h=4-(5-k)^2/4p$

Substitute into Eq. 2: $-4p(2+4-(5-k)^2/4p)=(11-k)^2$

Simplify: $-8p-16p+25-10k+k^2=121-22k+k^2$

Simplify: $-24p+25=121-12k$

Simplify: $p=0.5k-4$

From there, it should be relatively simple to find $h$ and $k$.

Answer 4

The equation of the parabola must be of form: x= a$y^2$ + by + c. With three points known to us, the following equations can be formed:

\begin{equation*} 25a + 5b + c = 4, \quad 121a + 11b + c = -2, \quad 441a + 21b + c = -4 \end{equation*}

Solving them gives

\begin{equation*} a=\frac{1}{20}, \quad b=-\frac{9}{5}, \quad c=\frac{47}{4} \end{equation*}

The y-intercepts can be found if we set x=0

\begin{equation*} \frac{y^2}{20}-\frac{9y}{5}+\frac{47}{4}=0 \implies y_1= 27.43,\quad y_2= 8.566 \end{equation*} These are the y-intercepts shown in the graph as well. Thus, the required equation of the parabola is

\begin{equation*} x= \frac{y^2}{20}-\frac{9y}{5}+\frac{47}{4} \end{equation*}